Chapter 3 Number Representation OBJECTIVES After reading this

Chapter 3 Number Representation

OBJECTIVES After reading this chapter, the reader should be able to : Convert a number from decimal to binary notation and vice versa. Understand the different representations of an integer inside a computer: unsigned, sign-and-magnitude, one’s complement, and two’s complement. Understand the Excess system that is used to store the exponential part of a floating-point number. Understand how floating numbers are stored inside a computer using the exponent and the mantissa.

3. 1 DECIMAL AND BINARY

Figure 3 -1 Decimal system

Figure 3 -2 Binary system

3. 2 CONVERSION

Figure 3 -3 Binary to decimal conversion

Example 1 Convert the binary number 10011 to decimal. Solution Write out the bits and their weights. Multiply the bit by its corresponding weight and record the result. At the end, add the results to get the decimal number. Binary Weights 1 0 0 1 1 16 8 4 2 1 --------------------16 + 0 + 2 + 1 Decimal 19

Example 2 Convert the decimal number 35 to binary. Solution Write out the number at the right corner. Divide the number continuously by 2 and write the quotient and the remainder. The quotients move to the left, and the remainder is recorded under each quotient. Stop when the quotient is zero. 0 1 2 4 8 17 35 Dec. Binary 1 0 0 0 1 1

Figure 3 -4 Decimal to binary conversion

3. 4 INTEGER REPRESENTATION

Figure 3 -5 Range of integers

Figure 3 -6 Taxonomy of integers

Unsigned integers format • Unsigned integer an integer without a sign. • Range: 0 … ( 2 N – 1 ) Table 3. 1 Range of unsigned integers # of Bits Range -----------------------8 0 255 16 0 65, 535

Example 3 Store 7 in an 8 -bit memory location. Solution 1. First change the number to binary 111. 2. Add five 0 s to make a total of N (8) bits, 00000111. The number is stored in the memory location.

Example 4 Store 258 in a 16 -bit memory location. Solution 1. First change the number to binary 100000010. 2. Add seven 0 s to make a total of N (16) bits, 000000010. The number is stored in the memory location.

Table 3. 2 Example of storing unsigned integers in two different computers Decimal ------7 234 258 24, 760 1, 245, 678 8 -bit allocation ------00000111 11101010 overflow 16 -bit allocation ---------------000000011101010 000000010 0110000010111000 overflow

overflow error l An error that occurs when the computer attempts to handle a number that is too large for it. l Every computer has a well-defined range of values that it can represent. If during execution of a program it arrives at a number outside this range, it will experience an overflow error.

Example 5 Interpret 00101011 in decimal if the number was stored as an unsigned integer. Solution Using the procedure shown in Figure 3. 3 , the number in decimal is 43.

Sign-and-magnitude format S b. N-2 … … … … b 1 b 0 • leftmost bit – represent the sign bit S Ø 0 for positive Ø 1 for negative • Range: -( 2 N-1 – 1 ) … +( 2 N-1 – 1 )

Note: There are two 0 s in sign-and-magnitude representation: positive and negative. In an 8 -bit allocation: +0 0000 -0 10000000

Table 3. 3 Range of sign-and-magnitude integers # of Bits -----8 16 32 Range ----------------------------127 -0 +0 +127 -32767 -0 +0 +32767 -2, 147, 483, 647 -0 +0 +2, 147, 483, 647

Note: In sign-and-magnitude representation, the leftmost bit defines the sign of the number. If it is 0, the number is positive. If it is 1, the number is negative.

Example 6 Store +7 in an 8 -bit memory location using sign-and-magnitude representation. Solution First change the number to binary 111. Add four 0 s to make a total of N-1 (7) bits, 0000111. Add an extra zero because the number is positive. The result is: 00000111

Example 7 Store – 258 in a 16 -bit memory location using sign-and-magnitude representation. Solution First change the number to binary 100000010. Add six 0 s to make a total of N-1 (15) bits, 00000010. Add an extra 1 because the number is negative. The result is: 100000010

Table 3. 4 Example of storing sign-and-magnitude integers in two computers Decimal ------+7 -124 +258 -24, 760 8 -bit allocation ------00000111 11111100 overflow 16 -bit allocation ---------------0000000111 100001111100 000000010 1110000010111000

Example 8 Interpret 1011 in decimal if the number was stored as a sign-and-magnitude integer. Solution Ignoring the leftmost bit, the remaining bits are 0111011. This number in decimal is 59. The leftmost bit is 1, so the number is – 59.

one’s complement format • to represent a positive number Ø unsigned integer • to represent a negative number Ø complete the positive number • Range: -( 2 N-1 – 1 ) … +( 2 N-1 – 1 )

Note: There are two 0 s in one’s complement representation: positive and negative. In an 8 -bit allocation: +0 0000 -0 1111

Table 3. 5 Range of one’s complement integers # of Bits ----8 16 32 Range ----------------------------127 -0 +0 +127 -32767 -0 +0 +32767 -2, 147, 483, 647 -0 +0 +2, 147, 483, 647

Note: In one’s complement representation, the leftmost bit defines the sign of the number. If it is 0, the number is positive. If it is 1, the number is negative.

Example 9 Store +7 in an 8 -bit memory location using one’s complement representation. Solution First change the number to binary 111. Add five 0 s to make a total of N (8) bits, 00000111. The sign is positive, so no more action is needed. The result is: 00000111

Example 10 Store – 258 in a 16 -bit memory location using one’s complement representation. Solution First change the number to binary 100000010. Add seven 0 s to make a total of N (16) bits, 000000010. The sign is negative, so each bit is complemented. The result is: 111111101

Table 3. 6 Example of storing one’s complement integers in two different computers Decimal ------+7 -7 +124 -124 +24, 760 -24, 760 8 -bit allocation ------00000111 11111000 01111100 10000011 overflow 16 -bit allocation ---------------0000000111 1111111000 00000111110000011 0110000010111000 1001111101000111

Example 11 Interpret 11110110 in decimal if the number was stored as a one’s complement integer. Solution 1. The leftmost bit is 1, so the number is negative. 2. First complement it. The result is 00001001. 3. The complement in decimal is 9. 4. So the original number was – 9. • Note that complement of a complement is the

Note: • One’s complement means reversing all bits. • If you one’s complement a positive number, you get the corresponding negative number. • If you one’s complement a negative number, you get the corresponding positive number. • If you one’s complement a number twice, you get the original number.

two’s complement format • Storing two’s complement : 1. The number is change to binary 2. 0 s are added to the left so that there is a total of N bits. 3. If the sign is negative, leave all the rightmost 0 s and the first 1 unchanged. Complete the rest of the bits. • Range: -( 2 N-1) … +( 2 N-1 – 1 )

Note: Two’s complement is the most common, the most important, and the most widely used representation of integers today.

Table 3. 7 Range of two’s complement integers # of Bits ----8 16 32 Range ----------------------------128 0 +127 -32, 768 0 +32, 767 -2, 147, 483, 648 0 +2, 147, 483, 647

Note: In two’s complement representation, the leftmost bit defines the sign of the number. If it is 0, the number is positive. If it is 1, the number is negative.

Example 12 Store +7 in an 8 -bit memory location using two’s complement representation. Solution First change the number to binary 111. Add five 0 s to make a total of N (8) bits, 00000111. The sign is positive, so no more action is needed. The result is: 00000111

Example 13 Store – 40 in a 16 -bit memory location using two’s complement representation. Solution 1. First change the number to binary 101000. 2. Add ten 0 s to make a total of N (16) bits, 00000101000. 3. The sign is negative, so leave the rightmost 0 s up to the first 1 (including the 1) unchanged and complement the rest. The result is: 11111011000

Table 3. 8 Example of storing two’s complement integers in two different computers Decimal ------+7 -7 +124 -124 +24, 760 -24, 760 8 -bit allocation ------00000111 11111001 01111100 10000100 overflow 16 -bit allocation ---------------0000000111 1111111001 00000111110000100 0110000010111000 1001111101001000

Note: There is only one 0 in two’s complement: In an 8 -bit allocation: 0 0000

Example 14 Interpret 11110110 in decimal if the number was stored as a two’s complement integer. Solution 1. The leftmost bit is 1. The number is negative. 2. Leave 10 at the right alone and complement the rest. The result is 00001010. 3. The two’s complement number is 10. 4. So the original number was – 10.

Note: • Two’s complement can be achieved by reversing all bits except the rightmost bits up to the first 1 (inclusive). • If you two’s complement a positive number, you get the corresponding negative number. • If you two’s complement a negative number, you get the corresponding positive number. • If you two’s complement a number twice, you get the original number.

Table 3. 9 Summary of integer representation Contents of Memory ------0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 Unsigned ------0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Sign-and. Magnitude ----+0 +1 +2 +3 +4 +5 +6 +7 -0 -1 -2 -3 -4 -5 -6 -7 One’s Complement ----+0 +1 +2 +3 +4 +5 +6 +7 -7 -6 -5 -4 -3 -2 -1 -0 Two’s Complement -------+0 +1 +2 +3 +4 +5 +6 +7 -8 -7 -6 -5 -4 -3 -2 -1

3. 5 EXCESS SYSTEM

Excess system • Magic number Ø A positive number used in the conversion process Ø Normally 2 N-1 or 2 N-1 – 1 • Representation 1. Add the positive number 2. Change to binary • Interpretation 1. Change to decimal 2. Subtract the positive number Original integer +M -M Excess-M

Example 15 Represent – 25 in Excess_127 using an 8 -bit allocation. Solution First add 127 to get 102. This number in binary is 1100110. Add one bit to make it 8 bits in length. The representation is 0110.

Example 16 Interpret 11111110 if the representation is Excess_127. Solution First change the number to decimal. It is 254. Then subtract 127 from the number. The result is decimal 127.

3. 5 FLOATING-POINT REPRESENTATION

Floating-Point Representation • Floating-Point number Ø Integer Ø fraction • Convert a floating-point number to binary 1. Convert the integer to binary 2. Convert the fraction to binary 3. Put a decimal point between the two parts.

Figure 3 -7 Changing fractions to binary

Example 17 Transform the fraction 0. 875 to binary Solution Write the fraction at the left corner. Multiply the number continuously by 2 and extract the integer part as the binary digit. Stop when the number is 0. 0. 0. 875 1. 750 1. 5 0 . 1 1 1. 0 0. 0 1

Example 18 Transform the fraction 0. 4 to a binary of 6 bits. Solution Write the fraction at the left cornet. Multiply the number continuously by 2 and extract the integer part as the binary digit. You can never get the exact binary representation. Stop when you have 6 bits. 0. 4 0. 8 1. 6 0 . 0 1 1. 2 0. 4 0. 8 1. 6 1 0 0 1

Normalization • We need a standard representation of floating-point number • Normalization Ø the moving of the decimal point so that there is only one 1 to the left of the decimal point. Ø 1. xxxxxx Ø Multiply 2 e, where e is the number of bits that the decimal point moved v positive for left movement v negative for right movement

Table 3. 10 Example of normalization Original. Number Move ----------------- 6 +1010001. 1101 +26 2 -111. 000011 -22 6 +0. 00000111001 +2 -6 3 -001110011 -0. 001110011 -2 -3 Normalized ------x 1. 01000111001 x 1. 11000011 x 1. 110011

Normalization • A normalized number Ø Sign Ø Exponent § § § The movement of decimal point Excess representation # of bits - define the range Ø Mantissa § § Binary number to the right of decimal point # of bits - define the precision • Note: the 1 to the left of decimal point is not stored; it is understood. • + 26 x 1. 0001110101 Ø Sign : + Ø Exponent : 6 Ø Mantissa : 0001110101

Figure 3 -8 IEEE standards for floating-point representation

Example 19 Show the representation of the normalized number + 26 x 1. 01000111001 Solution • The sign is positive. • The Excess_127 representation of the exponent is 133. • You add extra 0 s on the right to make it 23 bits. The number in memory is stored as: 0 10000101 01000111001000000

Table 3. 11 Example of floating-point representation -22 +2 -6 -2 -3 Number ------x 1. 11000011 x 1. 110011 Sign ---1 0 1 Exponent -----10000001 01111100 Mantissa ---------------110000000000 11001000000000 110000000000

Example 20 Interpret the following 32 -bit floating-point number 1 0111110000000000 Solution • The sign is negative. • The exponent is – 3 (124 – 127). • The number after normalization is -2 -3 x 1. 110011

3. 6 HEXADECIMAL NOTATION

• 81. 5625 • +1010001. 11001 • +26 x 1. 01000111001 • 0 10000101 01000111001000000 • x 42 A 39000